A008957 Triangle of central factorial numbers T(2*n,2*n-2*k), k >= 0, n >= 1 (in Riordan's notation).
1, 1, 1, 1, 5, 1, 1, 14, 21, 1, 1, 30, 147, 85, 1, 1, 55, 627, 1408, 341, 1, 1, 91, 2002, 11440, 13013, 1365, 1, 1, 140, 5278, 61490, 196053, 118482, 5461, 1, 1, 204, 12138, 251498, 1733303, 3255330, 1071799, 21845, 1, 1, 285, 25194, 846260, 10787231
Offset: 1
Examples
The triangle starts: 1; 1, 1; 1, 5, 1; 1, 14, 21, 1; 1, 30, 147, 85, 1; 1, 55, 627, 1408, 341, 1; 1, 91, 2002, 11440, 13013, 1365, 1;
References
- J. Riordan, Combinatorial Identities, Wiley, 1968, p. 217, Table 6.2(a).
- R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see Problem 5.8.
Links
- Reinhard Zumkeller, Rows n = 1..100 of triangle, flattened
- F. Alayont and N. Krzywonos, Rook Polynomials in Three and Higher Dimensions, 2012. [From _N. J. A. Sloane_, Jan 02 2013]
- D. Dumont, Interprétations combinatoires des nombres de Genocchi, Duke Math. J., 41 (1974), 305-318.
- D. E. Knuth, Johann Faulhaber and Sums of Powers, arXiv:math/9207222, Jul 1992. See bottom of page 10. [From _Michael Somos_, May 08 2018]
- Petro Kolosov, Polynomial identities involving central factorial numbers, GitHub, 2024. See p. 6.
Programs
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Haskell
a008957 n k = a008957_tabl !! (n-1) (k-1) a008957_row n = a008957_tabl !! (n-1) a008957_tabl = map reverse a036969_tabl -- Reinhard Zumkeller, Feb 18 2013
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Maple
A036969 := proc(n,k) local j; 2*add(j^(2*n)*(-1)^(k-j)/((k-j)!*(k+j)!),j=1..k); end; # Gives rows of triangle in reversed order
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Mathematica
t[n_, n_] = t[n_, 1] = 1; t[n_, k_] := t[n-1, k-1] + k^2 t[n-1, k]; Flatten[Table[t[n, k], {n, 1, 10}, {k, n, 1, -1}]][[1 ;; 50]] (* Jean-François Alcover, Jun 16 2011 *)
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PARI
{T(n, k) = if( n<1 || k>n, 0, n==k || k==1, 1, T(n-1, k-1) + k^2 * T(n-1, k))}; \\ Michael Somos, May 08 2018
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Sage
def A008957(n, k): m = n - k return 2*sum((-1)^(j+m)*(j+1)^(2*n)/(factorial(j+m+2)*factorial(m-j)) for j in (0..m)) for n in (1..7): print([A008957(n, k) for k in (1..n)]) # Peter Luschny, May 10 2018
Formula
From Michael Somos, May 08 2018: (Start)
T(n, k) = T(n-1, k-1) + k^2 * T(n-1, k), where T(n, n) = T(n, 1) = 1.
E.g.f.: x^2 * (cosh(sinh(y*x/2) / (x/2)) - 1) = (1*x^2)*y^2/2! + (1*x^2 + 1*x^4)*y^4/4! + (1*x^2 + 5*x^4 + 1*x^6)*y^6/6! + (1*x^2 + 14*x^4 + 21*x^6 + 1*x^8)*y^8/8! + ... (End)
Extensions
More terms from Vladeta Jovovic, Apr 16 2000
Comments